Precalculus (6th Edition) Blitzer

a.)-$\sqrt 3$ b.)$\sqrt 3$
a.)tan$\frac{5\pi}{3}$ corresponds to $\frac{y}{x}$ value. tan$\frac{5\pi}{3}$ = $\frac{-\frac{\sqrt 3}{2}}{\frac{1}{2}}$ = -$\sqrt 3$ b.)Tangent is an odd function, so tan(-t) = -tant. tan-$\frac{5\pi}{3}$ = - tan$\frac{5\pi}{3}$ = $\sqrt 3$